Understanding why $V^B$ is full In Chapter 14 of Jech's Set Theory, he introduces the notion of the Boolean-valued model $V^B$ where $B$ is a complete Boolean algebra (see here for the definition of a Boolean algebra, and completeness just means it satisfies the supremum property where supremums are defined as you'd expect them to be).

We shall define $V^B$ as a generalization of $V$: Instead of (two-valued) sets, we consider "Boolean-valued" sets, i.e. functions that assign Boolean values to their "elements." Thus we define $V^B$ as follows:
$$(\text{i})\quad V_0^B=\varnothing\\
(\text{ii})\quad V_{\alpha+1}^B=\text{the set of all functions $x$ with $\text{dom}(x)\subseteq V_\alpha^B$ and values in $B$}\\
V_\alpha^B=\bigcup_{\beta<\alpha}V_\beta^B \text{ if $\alpha$ is a limit ordinal}\\
(\text{iii})\quad V^B = \bigcup_{\alpha\in\text{Ord}} V_\alpha^B.$$

(side note: is there a way to do enumerate or align environments in MathJax?)

The definition of $||x\in y||$ and $||x=y||$ is motivated by (14.14)  and the requirement that $x(t)\leq ||t\in x||$. We define Boolean values by induction. Each $x\in V^B$ is assigned the rank in $V^B$,
$$\rho(x)=\text{the least $\alpha$ such that $x\in V_{\alpha+1}^\beta$}.$$
The forthcoming definition is by induction on pairs $(\rho(x),\rho(y))$, under the canonical well-ordering. To make the notation more suggestive, we introduce the following Boolean operation that corresponds to the implication:
$$u\Rightarrow v=-u+v.$$
Let
$$(\text{i}) \quad ||x\in y||=\sum_{t\in\text{dom }y} (||x=t||\cdot y(t)),\\
(\text{ii}) \quad ||x\subseteq y||=\sum_{t\in\text{dom }x} (x(t)\Rightarrow||t\in y||), \text{ and}\\
(\text{iii})\quad ||x=y||=||x\subseteq y||\cdot ||y\subseteq x||.$$

My current point of confusion is the proof that $V^B$ is full, Lemmas 14.18 and 14.19.

Lemma 14.18. If $W$ is a set of pairwise disjoint elements of $B$ and if $a_u,u\in W$, are elements of $V^B$, then there exists some $a\in V^B$ such that $u\leq ||a=a_u||$ for all $u\in W$.
Proof. Let $D=\bigcup_{u\in W} \text{dom}(a_u)$, and for every $t\in D$, let $a(t)=\sum\{u\cdot a_u(t):u\in W\}$. Since the $u$'s are pairwise disjoint, we have $u\cdot a(t)=u\cdot a_u(t)$ for each $u\in W$ and each $t\in D$. In other words, $u\leq (a(t)\Rightarrow a_u(t))$ and $u\leq (a_u(t)\Rightarrow a(t))$, and so $u\leq ||a=a_u||$.
Lemma 14.19. $V^B$ is full. Given a formula $\varphi(x,\dots)$, there exists some $a\in V^B$ such that (14.10) holds, i.e.
$$||\varphi(a,\dots)||=||\exists x\varphi (x,...)||.$$
Proof. In (14.10), $\leq$ holds for every $a$. We wish to find an $a\in V^B$ such that $\geq$ holds. Let $u_0=||\exists x\varphi(x,\dots)||$. Let
$$D=\{u\in B:\text{ there is some $a_u$ such that $u\leq ||\varphi(a_u,\dots)||$}\}.$$
It is clear that $D$ is open and dense below $u_0$. Let $W$ be a maximal set of pairwise disjoint elements of $D$; clearly, $\sum\{u:u\in W\}\geq u_0$. By Lemma 14.18 there exists some $a\in V^B$ such that $u\leq ||a=a_u||$ for all $u\in W$. Thus for each $u\in W$ we have $u\leq ||\varphi(a,\dots)||$, and hence $u_0\leq ||\varphi(a,\dots)||$.

I'm stuck on the following questions:

*

*In the proof of 14.18, how does it follow from $u\cdot a(t)=u\cdot a_u(t)$ for all $u,t$ that $u\leq (a(t)\Rightarrow a_u(t))$ and $u\leq (a_u(t)\Rightarrow a(t))$? Once we have this, how does it follow that $u\leq ||a=a_u||$?


*Why is it the case that $D$ is open and dense below $u_0$? I'm having a lot of trouble conceptualizing what the set $D$ is in the first place, particularly since there don't seem to be any restrictions on $a_u$ so I don't see why it's not all of $B$.


*Once we apply Lemma 14.18 and get such an $a$, how does it follow that $u\leq ||\varphi(a,\dots)||$, and how does it follow from that that $u_0\leq ||\varphi(a,\dots)||$?
 A: If $u\cdot v= u \cdot w,$ then $$u\cdot( v \Rightarrow w) = u\cdot(-v+w) = u\cdot (-v+v) = u,$$ so $u\le v\Rightarrow w.$ Since this and a couple other questions of yours are just matters of Boolean algebra "arithmetic" and definitions, I would advise carefully going over some exercises on this. Perhaps more importantly, you should work on intuition here: this identity can be seen by drawing a Venn diagram or using reasoning from elementrary set theory (Boolean algebras obey the same abstract rules as set algebras and $\le$ is analogous to $\subseteq$. This is a rare opportunity to actually visualize something... don't waste it!).
For the second part of this question, I'm afraid Jech has made a slight error here, since $a_u(t)$ is not necessarily defined for all $t\in D.$ One can instead just define $a(t) = \sum_u u\cdot \Vert t\in a_u\Vert.$ This unfortunately throws off the rhythm of the proof, but we can start from scratch. We want to show that a) for all $t\in D,$ and $u\in W,$ $ u\le a(t)\Rightarrow \Vert t\in a_u\Vert$ and b) for all $u\in W$ and $t\in\operatorname{dom}(a_u),$ $u\le a_u(t) \Rightarrow\Vert t\in a\Vert.$ (Since this is how to show $u\le \Vert a=a_u\Vert,$ per the definition 14.16 iii.)
For part a, since $W$ is an antichain, $u\cdot a(t) = \Vert t\in a_u\Vert$ which immediately implies $u\le a(t)\Rightarrow \Vert t\in a_u\Vert$ (why?). For part b, $$ u\cdot a_u(t) \le u\cdot \Vert t\in a_u\Vert\le a(t)\le \Vert t\in a\Vert.$$

We have $u\in D$ if and only if there is some $a\in V^B$ such that $u\le \Vert \varphi(a)\Vert$ (actually, we should probably also exclude $u=0$ for good order... more on that later). When would every $u\in B$ have that property? Exactly when $1$ has that property, i.e. when there is an $a\in V^B$ with $\Vert \varphi(a) \Vert = 1.$ There's no reason why this should need to be the case. In particular, it implies $\Vert \exists x \varphi(x) \Vert = 1$, and why should that be true?
Jech is making another slight oversight here (this one more of an omitted detail than an error/typo). We should be separating out the case when $\Vert \exists x \varphi(x) \Vert = 0$... the result is trivially true in that case since then any $a$ has $\Vert \varphi(a) \Vert = 0,$ but it's important to assume $u_0\ne 0$ in what follows. Also, as I mentioned before, $D$ should really be $\{u\in B^+:\exists a\in V^B\; u\le \Vert \varphi(a)\Vert \},$ since when we talk about dense sets and the like we are talking about subsets of $B^+.$
$D$ is obviously open (though I don't think that's actually relevant to what follows). If it weren't dense below $u_0$, there would be a $v\le u_0$ such that for all $w\le v$ with $w\ne 0,$ and all $a\in V^B,$ $w\nleq \Vert \varphi(a) \Vert.$ Another way of saying this last part is that for all $a\in V^B,$ $v\cdot \Vert \varphi(a) \Vert = 0$ (there are no common lower bounds of $v$ and $\Vert \varphi(a) \Vert$ other than $0$). Another way of saying this is that for all $a\in V^B,$ $v\le -\Vert \varphi(a)\Vert,$ which means $v\le \prod_a -\Vert \varphi(a)\Vert= - \Vert \exists x \varphi(x)\Vert.$ But this is impossible, since $v\le u_0\le \Vert \exists\varphi(x)\Vert.$

We have $u \le \Vert \varphi(a_u) \Vert$ and $u\le \Vert a = a_u\Vert,$ so $$ u\le \Vert \varphi(a_u) \Vert\cdot \Vert a = a_u\Vert \le \Vert \varphi(a) \Vert.$$ Jech already mentioned that $\sum W \ge u_0$ (actually, $\sum W = u_0$) from which it is clear that $u_0\le \sum W \le \Vert \varphi(a) \Vert.$ As for why $\sum W = u_0,$ note each $u\in W$ has $u\le u_0,$ and also that $W$ is a maximal antichain in the dense-below-$u_0$ set $D.$ If it weren't the case then let $z = u_0 - \sum W\ne 0,$ and $z\le u_0$ so there is a $z'\ne 0$ with $z'\le z$ and $z'\in D.$ Then $z'\in D,$ but $z'\cdot u=0$ for all $u\in W$ (since $z'\le -\sum W$ ), so this contradicts the maximality of $W.$ (Try to build intuition for this last argument by visualizing it in terms of set-algebras... it's important.)
